Sequences and Series Test

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Sequences and Series Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Value of $9 + 99 + 999 + ....$ upto $n$ terms is

A
$\displaystyle \frac{10^{n}-9n-10}{9}$

B
$\displaystyle \frac{10^{n}-9n-10}{81}$

C
$\displaystyle \frac{10^{n+1}-9n-10}{81}$

D
$\displaystyle \frac{10^{n+1}-9n-10}{9}$

Solution

Given, $9 + 99 + 999 + ....$ upto $n$ terms
we can write this series in this form
   $s=(10-1)+(100-1)+(1000-1).......$ up to $n$ terms.
    $= [(10+100+1000+......)+(-1-1-1-.......\text {up to n terms})]$
    $= [(10+10^2+10^3+.....\text {up to n terms})-n]$                 ........(i)
sum of n terms of GP. $=\dfrac {a(r^n-1)}{r-1}$ , where $a$ is first term and $r$ is the common ratio of the GP.
$\text {We have the GP } 10+10^2+10^3+.... \text {up to n terms}$ , here first term $a=10$ and common ratio $r=10$
So, $10+10^2+10^3+...... \text {n terms}= \dfrac {10(10^n-1)}{10-1} = \dfrac {10^{n+1}-10}{9}$ On Substituting $s=10+10^2+10^3+...... \text {n terms}= \dfrac {10^{n+1}-10}{9}$ in equation (i). we get,
    $=\dfrac {10^{n+1}-10}{9}-n = \dfrac {10^{n+1}-9n-10}{9}$
$\Rightarrow 9+99+999+.......\text {up to n terms} = \dfrac {10^{n+1}-9n-10}{9}$
Option D is correct.
Question 2
1 / -0

The sum of the series $4 + 8 + 16 + 32 + .......$ . till $10$ terms is

A
$6\, \times\, 2^{10}\, -\, 1$

B
$4\, \times\, 2^{10}\, +\, 1$

C
$4(2^{10}\, -\, 1)$

D
$4(2^{10}\, +\, 1)$

Solution

Let $S_{10}=4 + 8 + 16 + 32 + ........$ upto $10$ terms
$=4(1+2+4+8+.........$ .upto $10$ terms
Clearly, $1+2+2^2+2^3+...$ is an GP with first term $=1$ and common ratio $=2$ .
$\therefore S_{10}=4[\displaystyle \frac{1(2^{10}-1)}{2-1}]=4(2^{10}-1)$
Thus the sum of the given series is $4(2^{10}-1)$
Question 3
1 / -0

$\displaystyle \sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } } -i\cos { \frac { 2q\pi }{ 11 } } \right) } \right] }^{ p } } =$

A
$8\left( 1-i \right)$

B
$16\left( 1-i \right)$

C
$48\left( 1-i \right)$

D
None of these

Solution

$\displaystyle\sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } -i } \sin { \frac { 2q\pi }{ 11 } } \right) }$
$\displaystyle=\left\{ \sin { \frac { 2\pi }{ 11 } } +\sin { \frac { 4\pi }{ 11 } } +...+10 \right\}- i\left\{ \cos { \frac { 2\pi }{ 11 } } +\cos { \frac { 4\pi }{ 11 } } +...+10 \right\}$
$\displaystyle=\frac { \sin { \left( \frac { 2\pi }{ 11 } +\frac { 9\pi }{ 11 } \right) } \sin { \frac { 10\pi }{ 11 } } }{ \sin { \frac { \pi }{ 11 } } } -i\frac { \cos { \left( \frac { 2\pi }{ 11 } +\frac { 9\pi }{ 11 } \right) \sin { \frac { 10\pi }{ 11 } } } }{ \sin { \frac { \pi }{ 11 } } }$ $=0-i\left( -1 \right) =i$ ...(1)
$\displaystyle\therefore S=\sum _{ p=1 }^{ 32 }{ \left( 3q+2 \right) { \left[ \sum _{ q=1 }^{ 10 }{ \left( \sin { \frac { 2q\pi }{ 11 } -i } \cos { \frac { 2q\pi }{ 11 } } \right) } \right] }^{ p } }$
$\displaystyle=\sum _{ p=1 }^{ 32 }{ \left( 3p+2 \right) { i }^{ p } } =3\sum _{ p=1 }^{ 32 }{ p } { i }^{ p }+2\sum _{ p=1 }^{ 32 }{ { i }^{ p } } =3A+2B$ ...(2)
Now, $\displaystyle A=i+2{ i }^{ 2 }+3{ i }^{ 3 }+...+32{ i }^{ 32 }$ $\displaystyle \Rightarrow Ai={ i }^{ 2 }+2{ i }^{ 3 }+...+31{ i }^{ 32 }+32{ i }^{ 33 }$
$\displaystyle\Rightarrow A\left( 1-i \right) =i+{ i }^{ 2 }+...+{ i }^{ 32 }-32{ i }^{ 33 }$ $\displaystyle =\frac { { i }^{ 32 }-1 }{ i-1 } -32i=-32i\quad \left[ \because { i }^{ 32 }=1 \right]$
$\displaystyle\therefore A=\frac { -32i }{ 1-i } =\frac { 32\left( 1+i \right) }{ 2 } =16\left( 1-i \right)$ ...(3)
and, $\displaystyle B=i+{ i }^{ 2 }+...+{ i }^{ 32 }=0$ ...(4)
Hence, $\displaystyle S=3\times 16\left( 1-i \right) =48\left( 1-i \right) .$
Question 4
1 / -0

Find the sum the infinite G.P.:
$1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$

A
$\displaystyle \frac{3}{5}$

B
$\displaystyle \frac{3}{2}$

C
$\displaystyle \frac{49}{27}$

D
$\displaystyle \frac{8}{5}$

Solution

Formula used: Sum of infinite GP
$S_\infty=a+ar+ar^2+ar^3+...=\dfrac{a}{1-r}$ , where $|r|<1$
Given, $1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}$
Clearly, it is a GP with first term $a=1$ and common difference $d=\frac13$
$\therefore1\, +\, \displaystyle {\frac{1}{3}\, +\, \frac{1}{9}\, +\, \frac{1}{27}\, +\, .......}=\frac{1}{1-\frac13}=\frac32$
Hence, option B is correct.
Question 5
1 / -0

Evaluate $7 + 77 + 777 + .............$ upto $n$ terms.

A
$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n]$

B
$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n\, -\, 10]$

C
$\displaystyle \frac{7}{81}[10^n\, -\, 9n\, -\, 10]$

D
$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, n\, -\, 10]$

Solution

Let $S=7+77+777+........+$ up to $n$ terms
$\Rightarrow \dfrac{7}{9}\left [ 9+99+999+..... \right ]$

$\Rightarrow \dfrac{7}{9}\left [\left (10-1 \right )+\left (10^2-1 \right )+\left (10^3+1 \right )+........up\quad to\quad n\quad terms \right ]$

$\Rightarrow \dfrac{7}{9}\left [10+10^2+10^3+......+10^n-n \right ]$

$\Rightarrow \dfrac{7}{9}\left [ 10\frac{10^{n}-1}{9}-n \right ]$

$\Rightarrow \dfrac{7}{81}\left [ 10^{n+1}-9n-10 \right ]$
Question 6
1 / -0

For $\frac {2^2+4^2+6^2+...+(2n)^2}{1^2+3^2+5^2+...+(2n-1)^2}$ to exceed 1.01, the maximum value of n is

A
99

B
100

C
101

D
150

Solution

$\frac {2^2[1^2+2^2+...+n^2]}{[1^2+3^2+5^2+...+(2n-1)^2]}$
$1^2+2^2+3^2+....+(2n)^2=\frac {2n(2n+1)(4n+1)}{6}$
$[1^2+3^2....+(2n-1)^2+2^2[1^2+2^n.....+n^2]$
$=\frac {2n(2n+1)(4n+1)}{6}$
$S+4\frac {n(n+1)(2nn+1)}{6}=\frac {2n(2n+1)(4n+1)}{6}$
$S+\frac {2n(2n+1)(4n+1)}{6}-\frac {4n(n+1)(2n+1)}{6}$
$=2n\left (\frac {2n+1}{6}\right )[4n+1-2n-2]$
$=\frac {2n(2n+1)(2n-1)}{6}$
$Ratio =\frac {4n(n+1)(2n+1)}{6}\times \frac {6}{2n(2n+1)(2n-1)}=\frac {2n+2}{2n-1}$
$\frac {2n+2}{2n-1} > \frac {101}{100}$
$200n+200 > 202 n-101$
$2n < 301$
$n < \frac {301}{2}\Rightarrow$ maximum value $=150$
Question 7
1 / -0

In the following question, the numbers/letters are arranged based on some pattern or principle.Choose the correct answer for the term marked by the symbol (?)

A
$18$

B
$40$

C
$21$

D
$24$

Solution
Question 8
1 / -0

The sum of 'n' terms of series $1^2+(1^2+2^2)+(1^2+ 2^2+ 3^2)+(1^2+2^2+ 3^2+4^2)+ .......$ will be

A
$\displaystyle \frac{1}{4} n (n + 1)^2 (n + 2)$

B
$\displaystyle \frac{1}{3} n (n + 1)^2 (n + 2)$

C
$\displaystyle \frac{1}{12} n (n + 1) (n + 2)^2$

D
$\displaystyle \frac{1}{12} n (n + 1)^2 (n + 2)$

Solution

The given series is ${ 1 }^{ 2 }+({ 1 }^{ 2 }+2^{ 2 })+({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 })+({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 }+4^{ 2 })+.....$

${ a }_{ n }=({ 1 }^{ 2 }+2^{ 2 }+3^{ 2 }+.......+n^{ 2 })$

$=\dfrac { n(n+1)(2n+1) }{ 6 } \\ =\dfrac { n(2n^{ 2 }+3n+1) }{ 6 } \\ =\dfrac { 2n^{ 3 }+3n^{ 2 }+n }{ 6 } \\ =\dfrac { 1 }{ 3 } n^{ 3 }+\dfrac { 1 }{ 2 } n^{ 2 }+\dfrac { 1 }{ 6 } n$

Therefore, we have:

${ S }_{ n }=\sum _{ k=1 }^{ n }{ { a }_{ k } } \\ =\sum _{ k=1 }^{ n }{ \left( \dfrac { 1 }{ 3 } k^{ 3 }+\dfrac { 1 }{ 2 } k^{ 2 }+\dfrac { 1 }{ 6 } k \right) } \\ =\dfrac { 1 }{ 3 } \sum _{ k=1 }^{ n }{ k^{ 3 } } +\dfrac { 1 }{ 2 } \sum _{ k=1 }^{ n }{ k^{ 2 } } +\dfrac { 1 }{ 6 } \sum _{ k=1 }^{ n }{ k } \\ =\dfrac { 1 }{ 3 } \times \dfrac { n^{ 2 }+(n+1)^{ 2 } }{ 2^{ 2 } } +\dfrac { 1 }{ 2 } \times \dfrac { n(n+1)(2n+1) }{ 6 } +\dfrac { 1 }{ 6 } \times \dfrac { n(n+1) }{ 2 }$
$=\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1) }{ 2 } +\dfrac { (2n+1) }{ 2 } +\dfrac { 1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n^{ 2 }+n+2n+1+1 }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { n(n+1)+2(n+1) }{ 2 } \right] \\ =\dfrac { n(n+1) }{ 6 } \left[ \dfrac { (n+1)(n+2) }{ 2 } \right] \\ =\dfrac { n(n+1)^{ 2 }(n+2) }{ 12 }$

Hence, the sum is $\dfrac { n(n+1)^{ 2 }(n+2) }{ 12 }$ .
Question 9
1 / -0

$11, 26, 56, 101, 161, .....$

A
$236$

B
$226$

C
$216$

D
$206$

Solution

Add $15, 30, 45, 60, 75$ to each number Next number is $161 + 75 = 236$
Question 10
1 / -0

The value of $\displaystyle\frac{1}{1\cdot2\cdot3}+\displaystyle\frac{1}{2\cdot3\cdot4}+\displaystyle\frac{1}{3\cdot4\cdot5}+\displaystyle\frac{1}{4\cdot5\cdot6}$ is equal to

A
$\displaystyle\frac{7}{30}$

B
$\displaystyle\frac{11}{30}$

C
$\displaystyle\frac{13}{30}$

D
$\displaystyle\frac{17}{30}$

Solution

Given $\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}$

$\Rightarrow \dfrac{1}{1\times 2\times3 }+\dfrac{1}{2\times 3\times 4}+\dfrac{1}{3\times 4\times 5}+\dfrac{1}{4\times 5\times 6}$

$\Rightarrow \dfrac{1}{6}+\dfrac{1}{24}+\dfrac{1}{60}+\dfrac{1}{120}$

$= \dfrac{20+5+2+1}{120}$

$= \dfrac{28}{120}$

$= \dfrac{7}{30}$

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Sequences and Series Test - 23

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Sequences and Series Test - 23

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Question 1

10n−9n−109\displaystyle \frac{10^{n}-9n-10}{9}910n−9n−10​

10n−9n−1081\displaystyle \frac{10^{n}-9n-10}{81}8110n−9n−10​

10n+1−9n−1081\displaystyle \frac{10^{n+1}-9n-10}{81}8110n+1−9n−10​

10n+1−9n−109\displaystyle \frac{10^{n+1}-9n-10}{9}910n+1−9n−10​

Solution

Question 2

6 × 210 − 16\, \times\, 2^{10}\, -\, 16×210−1

4 × 210 + 14\, \times\, 2^{10}\, +\, 14×210+1

4(210 − 1)4(2^{10}\, -\, 1)4(210−1)

4(210 + 1)4(2^{10}\, +\, 1)4(210+1)

Solution

Question 3

8(1−i)8\left( 1-i \right) 8(1−i)

16(1−i)16\left( 1-i \right) 16(1−i)

48(1−i)48\left( 1-i \right) 48(1−i)

None of these

Solution

Question 4

35\displaystyle \frac{3}{5}53​

32\displaystyle \frac{3}{2}23​

4927\displaystyle \frac{49}{27}2749​

85\displaystyle \frac{8}{5}58​

Solution

Question 5

781[10n + 1 − 9n]\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n]817​[10n+1−9n]

781[10n + 1 − 9n − 10]\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n\, -\, 10]817​[10n+1−9n−10]

781[10n − 9n − 10]\displaystyle \frac{7}{81}[10^n\, -\, 9n\, -\, 10]817​[10n−9n−10]

781[10n + 1 − n − 10]\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, n\, -\, 10]817​[10n+1−n−10]

Solution

Question 6

99

100

101

150

Solution

Question 7

181818

404040

212121

242424

Solution

Question 8

14n(n+1)2(n+2)\displaystyle \frac{1}{4} n (n + 1)^2 (n + 2)41​n(n+1)2(n+2)

13n(n+1)2(n+2)\displaystyle \frac{1}{3} n (n + 1)^2 (n + 2)31​n(n+1)2(n+2)

112n(n+1)(n+2)2\displaystyle \frac{1}{12} n (n + 1) (n + 2)^2121​n(n+1)(n+2)2

112n(n+1)2(n+2)\displaystyle \frac{1}{12} n (n + 1)^2 (n + 2)121​n(n+1)2(n+2)

Solution

Question 9

236236236

226226226

216216216

206206206

Solution

Question 10

730\displaystyle\frac{7}{30}307​

1130\displaystyle\frac{11}{30}3011​

1330\displaystyle\frac{13}{30}3013​

1730\displaystyle\frac{17}{30}3017​

Solution

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$\displaystyle \frac{10^{n}-9n-10}{9}$

$\displaystyle \frac{10^{n}-9n-10}{81}$

$\displaystyle \frac{10^{n+1}-9n-10}{81}$

$\displaystyle \frac{10^{n+1}-9n-10}{9}$

$6\, \times\, 2^{10}\, -\, 1$

$4\, \times\, 2^{10}\, +\, 1$

$4(2^{10}\, -\, 1)$

$4(2^{10}\, +\, 1)$

$8\left( 1-i \right)$

$16\left( 1-i \right)$

$48\left( 1-i \right)$

$\displaystyle \frac{3}{5}$

$\displaystyle \frac{3}{2}$

$\displaystyle \frac{49}{27}$

$\displaystyle \frac{8}{5}$

$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n]$

$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, 9n\, -\, 10]$

$\displaystyle \frac{7}{81}[10^n\, -\, 9n\, -\, 10]$

$\displaystyle \frac{7}{81}[10^{n\, +\, 1}\, -\, n\, -\, 10]$

$18$

$40$

$21$

$24$

$\displaystyle \frac{1}{4} n (n + 1)^2 (n + 2)$

$\displaystyle \frac{1}{3} n (n + 1)^2 (n + 2)$

$\displaystyle \frac{1}{12} n (n + 1) (n + 2)^2$

$\displaystyle \frac{1}{12} n (n + 1)^2 (n + 2)$

$236$

$226$

$216$

$206$

$\displaystyle\frac{7}{30}$

$\displaystyle\frac{11}{30}$

$\displaystyle\frac{13}{30}$

$\displaystyle\frac{17}{30}$